Unizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.

Monday, October 10, 2011

Construction problems in geometry are extremely important. They help to develop student's creativity, logical thinking and ability to combine analysis of a problem at hand with synthesis of existing methods into concrete solution. Solving problems is the most important purpose of UNIZOR.COM. Please examine these problems and try to solve them. The accompanying lecture contains all the solutions, but watch it only after you spent sufficient amount of time trying to solve these problems yourself.

1. Construct a sum of two angles.
In details, this construction problem has the following meaning, applicable to other problems as well (so, we will not repeat it).
Given two angles and a ray somewhere on a plane.
Using a given ray as one leg, construct another ray from the same starting point such that these two rays (one given and one constructed) form an angle, which has a measure equal to a sum of measures of two given angles.

2. Construct a difference between a bigger and a smaller angle.

3. Given a sum and a difference between two angles, construct these angles.

4. Divide an angle into 4, 8 and 16 congruent parts.

5. Construct a triangle by its two sides and an angle opposite to a bigger side.
Is it always a unique triangle? Analyse the number of solutions of this problem.

6. Construct a triangle by its two sides and an angle opposite to a smaller side.
Is it always a unique triangle? Analyse the number of solutions of this problem.

7. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two other sides.

8. Construct an isosceles triangle by a base (a side, generally speaking, non-congruent to two others) and one of two congruent angles (usually called angles at the base or base angles).

9. Construct an isosceles triangle by one of two congruent sides and an angle between congruent sides (usually called an angle at the vertex).

10. Construct an isosceles triangle by one of two congruent sides and a base angle it forms with a base (a side, generally speaking, non-congruent to two others).

11. Construct a right triangle by its two legs (i.e. sides, perpendicular to each other, sometimes called also catheti).

12. Construct a right triangle by one of its legs (also known as cathetus) and a hypotenuse.

13. Construct a right triangle by a cathetus and an acute angle it forms with a hypotenuse.

14. Construct an isosceles triangle by an altitude to a base and one of two congruent sides.

15. Construct an isosceles triangle by an altitude to a base and an angle opposite to a base (i.e. vertex angle).

16. Construct an isosceles triangle by a base and an altitude towards one of two congruent sides.

17. Construct a right triangle by a hypotenuse and an acute angle it forms with one of catheti.

18. Given an angle and a point inside it. Construct a straight line crossing this point and cutting congruent segments from sides of a given angle.

19. Construct two segments by their given sum and difference.

20. Divide a given segment into 4, 8 and 16 congruent parts.

21. Given a straight line and two points anywhere on a plane. Construct a point on a given line equidistant to two given points.

22. Construct a point equidistant to three vertices of a given triangle.

23. Given an angle and a straight line that crosses both its sides. Construct a point on this line equidistant to two sides of a given angle.

24. Construct a point equidistant to three sides of a given triangle.

25. Given a straight line PQ and two points M and N outside it, but on the same side from it. Construct a point X on the line PQ such that angles ∠PXM and ∠QXN are congruent.

26. Construct a right triangle by one of its catheti and a sum of hypotenuse and the other cathetus.

27. Construct a triangle by a side, an angle it makes with another side and a sum of two other sides.

28. Given an angle and two points, one on each side of an angle. Construct a point equidistant from both sides of an angle and, at the same time, equidistant from two given points on angle's sides.
Notice, that distances to angle's sides should be equal to each other, but not necessarily equal to distances to given points.

Finally, I've reached a point when I can start introducing problems in Geometry. Problems below are, mostly, from more than 100 years old textbook on geometry by a famous Russian math teacher Kiselev.
Solving problems is the most important purpose of our site UNIZOR.COM. Please examine these problems and try to solve them. The accompanying lecture contains all the solutions, but watch it only after you spent sufficient amount of time trying to solve these problems yourself.

1. Prove that in an isosceles triangle two medians to congruent sides are congruent.

2. Prove that in an isosceles triangle two angle bisectors to congruent sides are congruent.

3. Prove that in an isosceles triangle two altitudes to congruent sides are congruent.

4. Prove that segments of two perpendicular bisectors to two congruent sides of an isosceles triangle between midpoint of one side and crossing of its bisector with another side are congruent.
More precisely, consider an isosceles triangle ΔABC with congruent sides AB and BC. Draw perpendicular bisectors of its congruent sides through midpoint M of side AB and midpoint N of side BC. Point P is a crossing point of the perpendicular bisector of side AB with side BC. Point Q is a crossing point of the perpendicular bisector of side BC with side AB. Prove that segments MP and NQ are congruent.

5. Prove that a line, perpendicular to an angle bisector, cuts from two rays forming this angle congruent segments, assuming that an angle is less that 180 degrees.
In details, let AB and AC be two rays forming angle ∠BAC, let AM be a bisector of this angle and P - any point on this bisector. Draw a perpendicular to AM that croses it at point P. This same perpendicular crosses rays AB at point X and AC at point Y. Prove that segments AX and AY are congruent.

6. Prove that median AM of a triangle ΔABC from vertex A to an opposite side BC is equidistant from vertices B and C.
Notice, that distance from a point to a line is measured as a length of a perpendicular dropped from this point to a line.

7. Prove that the length of median AM of a triangle ΔABC from vertex A to an opposite side BC is less than half sum of lengths of sides AB and AC in lies in between.
Hint: extend the median beyond side BC by doubling its length to point D and consider triangle ΔABD.

8. Prove that sum of lengths of all three medians of a triangle is less than its perimeter, but greater than its halp-perimeter.
Hint: use the triangle inequality and the previous problem.

9. Prove that sum of lengths of two diagonals of a quadrangle is less than its perimeter, but greater than its half-perimeter.

10. Given an angle ∠PMQ and two points on each side: A and B on side MP and A' and B' on side MQ such that corresponding pairs of segments are congruent:MA is congruent to MA' and MB is congruent to MB'.
Prove that crossing segments AB' and A'B cross on a bisector of angle ∠PMQ.
Notice, that this theorem implies an easy way to construct an angle bisector.

11. Given a sraight line PQ and two pairs of points symmetrical relative to this line: points A and A' are symmetrical relative to PQ, points B and B' are symmetrics too relative to the same line PQ.
Prove that there exists a point M on line PQ equidistant from all four points A, A', B and B'.

12. Given a straight line PQ and two points A and B on the same side from it. Find a point M on line PQ such that the sum of lengths of segments AM and MB is minimal.

13. Given an accute angle ∠PMQ and point A inside it. Find point X on side MP of this angle and point Y on its other side MQ such that a perimeter of triangle ΔAXY is minimal.

About Me

Mission: Our need for a major change in basic education has become so obvious and so critical that citizen action is mandated.

Product: UNIZOR provides the framework for that change; a place for the gathering of what education should be; a place for delivering greater value at a fraction of the cost. It sets the path for eventual replacement of obsolete educational methodologies.

Key: Parents are the customers and the main party in interest. They receive and value the educational product, which is supposed to maximize the future of students under their supervision. They personally or through representatives actively manage the educational process.